NTA Ans. (3)Ans. Bonusf(x)=x2−4x+9(x+5)(x−3) Let g(x)=x2−4x+9D<0g(x)>0 for x∈R
∴[f(−5)=0f(3)=0].So, f(x) is many-one. again,yx2−4xy+9y=x2+2x−15x2(y−1)−2x(2y+1)+(9y+15)=0 for ∀x∈R⇒D≥0D=4(2y+1)2−4(y−1)(9y+15)≥05y2+2y+16≤0(5y−8)(y+2)≤0
y∈[−2,58] range Note : If function is defined from f:R⟶R then only correct answer is option (3)⇒ Bonus